section 2: three-phase power fundamentalsweb.engr.oregonstate.edu/~webbky/ese470_files/section...

ESE 470 – Energy Distribution Systems

SECTION 2: THREE-PHASE POWER FUNDAMENTALS

K. Webb ESE 470

AC Circuits & Phasors2

K. Webb ESE 470

3

AC Electrical Signals

AC electrical signals (voltages and currents) are sinusoidal Generated by rotating machinery

Sinusoidal voltage (or current):

𝑣𝑣 𝑡𝑡 = 𝑉𝑉𝑝𝑝 cos 𝜔𝜔𝑡𝑡 + 𝜙𝜙 (1)

This is a time-domain or instantaneous form expression

Characterized by three parameters Amplitude Frequency Phase

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Amplitude

𝑣𝑣 𝑡𝑡 = 𝑉𝑉𝑝𝑝 cos 𝜔𝜔𝑡𝑡 + 𝜙𝜙

𝑉𝑉𝑝𝑝 in the above expression is amplitude or peak voltage We typically characterize power-system voltages and

currents in terms of their root-mean-square (rms) values

𝑉𝑉𝑟𝑟𝑟𝑟𝑟𝑟 = 1𝑇𝑇 ∫0

𝑇𝑇 𝑣𝑣 𝑡𝑡 2𝑑𝑑𝑡𝑡12 (2)

A signal delivers the same power to a resistive load as a DC signal equal to its rms value

For sinusoids:𝑉𝑉𝑟𝑟𝑟𝑟𝑟𝑟 = 𝑉𝑉𝑝𝑝

2(3)

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Euler’s Identity

Euler’s identity allows us to express sinusoidal signals as complex exponentials

𝑒𝑒𝑗𝑗𝑗𝑗𝑗𝑗 = cos 𝜔𝜔𝑡𝑡 + 𝑗𝑗 sin 𝜔𝜔𝑡𝑡 (4)

so𝑒𝑒𝑗𝑗 𝑗𝑗𝑗𝑗+𝜙𝜙 = cos 𝜔𝜔𝑡𝑡 + 𝜙𝜙 + 𝑗𝑗 sin 𝜔𝜔𝑡𝑡 + 𝜙𝜙 (5)

and 𝑉𝑉𝑝𝑝 cos 𝜔𝜔𝑡𝑡 + 𝜙𝜙 = 𝑉𝑉𝑝𝑝𝑅𝑅𝑒𝑒 𝑒𝑒𝑗𝑗 𝑗𝑗𝑗𝑗+𝜙𝜙

𝑉𝑉𝑝𝑝 cos 𝜔𝜔𝑡𝑡 + 𝜙𝜙 = 2 𝑉𝑉𝑟𝑟𝑟𝑟𝑟𝑟𝑅𝑅𝑒𝑒 𝑒𝑒𝑗𝑗 𝑗𝑗𝑗𝑗+𝜙𝜙 (6)

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Phasor Representation

Phasor representation simplifies circuit analysis when dealing with sinusoidal signals Drop the time-harmonic (oscillatory) portion of the signal representation

Known and constant Represent with rms amplitude and phase only

For example, consider the time-domain voltage expression

𝑣𝑣 𝑡𝑡 = 2 𝑉𝑉𝑟𝑟𝑟𝑟𝑟𝑟 cos 𝜔𝜔𝑡𝑡 + 𝜙𝜙

The phasor representation, in exponential form, is

𝑽𝑽 = 𝑉𝑉𝑟𝑟𝑟𝑟𝑟𝑟𝑒𝑒𝑗𝑗𝜙𝜙

Can also express in polar or Cartesian form

𝑽𝑽 = 𝑉𝑉𝑟𝑟𝑟𝑟𝑟𝑟∠𝜙𝜙 = 𝑉𝑉𝑟𝑟𝑟𝑟𝑟𝑟 cos 𝜙𝜙 + 𝑗𝑗𝑉𝑉𝑟𝑟𝑟𝑟𝑟𝑟 sin 𝜙𝜙

In these notes bold type will be used to distinguish phasors We’ll always assume rms values for phasor magnitudes

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Phasors

Circuit analysis in the phasor domain is simplified Derivative and integrals become algebraic expressions

Consider the voltage across inductance and capacitance:

Think of a phasor as a vector in the complex plane Has magnitude and angle

Time Domain Phasor Domain

Capacitor𝑣𝑣 𝑡𝑡 =

1𝑐𝑐� 𝑖𝑖 𝑡𝑡 𝑑𝑑𝑡𝑡 𝑽𝑽 =

1𝑗𝑗𝜔𝜔𝑗𝑗

𝑰𝑰

Inductor 𝑣𝑣 𝑡𝑡 = 𝐿𝐿𝑑𝑑𝑖𝑖𝑑𝑑𝑡𝑡

𝑽𝑽 = 𝑗𝑗𝜔𝜔𝐿𝐿𝑰𝑰

Resistor 𝑣𝑣 𝑡𝑡 = 𝑖𝑖 𝑡𝑡 𝑅𝑅 𝑽𝑽 = 𝑰𝑰𝑅𝑅

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Phasors

In general, in the phasor domain

𝑽𝑽 = 𝑰𝑰𝑍𝑍 (7)

and

𝑰𝑰 =𝑽𝑽𝑍𝑍

Ohm’s law 𝑍𝑍 is a complex impedance

Not a phasor, but also expressed in exponential, polar, or Cartesian form

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Phasors - Example

Determine 𝑖𝑖 𝑡𝑡 and 𝑣𝑣𝐿𝐿 𝑡𝑡 for the following circuit, driven by a 120 𝑉𝑉𝑟𝑟𝑟𝑟𝑟𝑟, 60 𝐻𝐻𝐻𝐻 source

At 60 𝐻𝐻𝐻𝐻 the inductor impedance is𝑗𝑗𝑋𝑋𝐿𝐿 = 𝑗𝑗𝜔𝜔𝐿𝐿 = 𝑗𝑗𝑗𝑗𝑗 ⋅ 60 𝐻𝐻𝐻𝐻 ⋅ 5 𝑚𝑚𝐻𝐻 = 𝑗𝑗𝑗.88 Ω

The total impedance seen by the source is𝑍𝑍 = 𝑅𝑅 + 𝑗𝑗𝑋𝑋𝐿𝐿 = 2 + 𝑗𝑗𝑗.88 Ω

Converting to polar form𝑍𝑍 = 𝑍𝑍 ∠𝜃𝜃

𝑍𝑍 = 𝑅𝑅2 + 𝑋𝑋2 = 2.74 Ω

𝜃𝜃 = tan−1𝑋𝑋𝑅𝑅

= 43°

𝑍𝑍 = 2.74∠43° Ω

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Phasors – Example

The source voltage is𝑣𝑣 𝑡𝑡 = 2 ⋅ 𝑗𝑗0𝑉𝑉𝑐𝑐𝑉𝑉𝑉𝑉 𝑗𝑗𝑗 ⋅ 60𝐻𝐻𝐻𝐻 ⋅ 𝑡𝑡

The source voltage phasor is𝑽𝑽 = 𝑗𝑗0∠0° 𝑉𝑉

The current phasor is

𝑰𝑰 =𝑽𝑽𝑍𝑍

=𝑗𝑗0∠0° 𝑉𝑉

2.74∠43° Ω= 43.7∠ − 43° 𝐴𝐴

We can use the current phasor to determine the phasor for the voltage across the resistor

𝑽𝑽𝑳𝑳 = 𝑰𝑰𝑅𝑅 = 43.7∠ − 43° ⋅ 2Ω

𝑽𝑽𝑳𝑳 = 87.4∠ − 43° 𝑉𝑉

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Phasors – Example

We have phasor representations for desired quantities

𝑰𝑰 = 43.7∠ − 43° 𝐴𝐴𝑽𝑽𝑳𝑳 = 87.4∠ − 43° 𝑉𝑉

We can now convert these to their time-domain expressions

𝑖𝑖 𝑡𝑡 = 2 ⋅ 43.7 𝐴𝐴 ⋅ cos 𝑗𝑗𝑗 ⋅ 60𝐻𝐻𝐻𝐻 ⋅ 𝑡𝑡 − 43°

𝑣𝑣 𝑡𝑡 = 2 ⋅ 87.4 𝑉𝑉 ⋅ cos 𝑗𝑗𝑗 ⋅ 60𝐻𝐻𝐻𝐻 ⋅ 𝑡𝑡 − 43°

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Phasor Diagrams12

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Phasor Diagrams

Phasors are complex values Magnitude and phase Vectors in the complex plane Can represent graphically

Phasor diagram Graphical representation of phasors in a circuit KVL and Ohm’s law expressed graphically

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Phasor Diagram – Example 1

Source voltage is the reference phasor𝑽𝑽𝑆𝑆 = 𝑗𝑗0∠0° 𝑉𝑉

Its phasor diagram:

Ohm’s law gives the current

𝑰𝑰 =𝑽𝑽𝑆𝑆

2 + 𝑗𝑗𝑗 Ω= 42.𝑗∠ − 45° 𝐴𝐴

Adding to the phasor diagram:

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Ohm’s law gives the inductor voltage𝑽𝑽𝐿𝐿 = 𝐼𝐼 ⋅ 𝑗𝑗𝜔𝜔𝐿𝐿 = 42.𝑗∠ − 45° 𝐴𝐴 ⋅ 𝑗𝑗𝑗 Ω𝑽𝑽𝐿𝐿 = 85∠45° 𝑉𝑉

Finally, KVL gives 𝑽𝑽𝑅𝑅𝑽𝑽𝑅𝑅 = 𝑽𝑽𝑆𝑆 − 𝑽𝑽𝐿𝐿𝑽𝑽𝑅𝑅 = 𝑗𝑗0∠0° 𝑉𝑉 − 85∠45° 𝑉𝑉𝑽𝑽𝑅𝑅 = 85∠ − 45°

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Source voltage is the reference phasor

𝑽𝑽𝑆𝑆 = 2.4∠0° 𝑘𝑘𝑉𝑉

Ohm’s law gives the current

𝑰𝑰 =𝑽𝑽𝑆𝑆

3.5 + 𝑗𝑗3 Ω= 5𝑗𝑗∠ − 41° 𝐴𝐴

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Ohm’s law gives the resistor voltage𝑽𝑽𝑅𝑅 = 𝐼𝐼 ⋅ 𝑅𝑅𝑽𝑽𝑅𝑅 = 521∠ − 41° 𝐴𝐴 ⋅ 1.5 Ω𝑽𝑽𝑅𝑅 = 78𝑗∠ − 41° 𝑉𝑉

KVL gives 𝑽𝑽2𝑽𝑽2 = 𝑽𝑽𝑆𝑆 − 𝑽𝑽𝐿𝐿𝐿𝐿𝐿𝐿𝐿𝐿𝑅𝑅𝑽𝑽2 = 2.4∠0° 𝑘𝑘𝑉𝑉 − 78𝑗∠ − 41° 𝑉𝑉𝑽𝑽2 = 1.88∠𝑗5.7° 𝑘𝑘𝑉𝑉

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Drop across the inductor:𝑽𝑽𝐿𝐿𝐿𝐿𝐿𝐿𝐿𝐿𝐿𝐿 = 521∠ − 41° 𝐴𝐴 ⋅ 𝑗𝑗𝑗 Ω𝑽𝑽𝐿𝐿𝐿𝐿𝐿𝐿𝐿𝐿𝐿𝐿 = 1.04∠49° 𝑘𝑘𝑉𝑉

KVL gives the voltage across the load𝑽𝑽𝑅𝑅 = 𝑽𝑽2 − 𝑽𝑽𝐿𝐿𝐿𝐿𝐿𝐿𝐿𝐿𝐿𝐿𝑽𝑽𝑅𝑅 = 1.88∠𝑗5.7°𝑘𝑘𝑉𝑉 − 1.04∠49° 𝑘𝑘𝑉𝑉𝑽𝑽𝑅𝑅 = 1.𝑗6∠ − 14° 𝑘𝑘𝑉𝑉

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Alternatively, treat the line as a single impedance

𝑽𝑽𝐿𝐿𝐿𝐿𝐿𝐿𝐿𝐿 = 𝐼𝐼 ⋅ 𝑍𝑍𝐿𝐿𝐿𝐿𝐿𝐿𝐿𝐿𝑽𝑽𝐿𝐿𝐿𝐿𝐿𝐿𝐿𝐿 = 521∠ − 41° 𝐴𝐴 ⋅ 1.5 + 𝑗𝑗𝑗 Ω𝑽𝑽𝐿𝐿𝐿𝐿𝐿𝐿𝐿𝐿𝐿𝐿 = 1.3∠𝑗𝑗.5° 𝑘𝑘𝑉𝑉

KVL gives the voltage across the load𝑽𝑽𝑅𝑅 = 𝑽𝑽𝑆𝑆 − 𝑽𝑽𝐿𝐿𝐿𝐿𝐿𝐿𝐿𝐿𝑽𝑽𝑅𝑅 = 2.4∠0°𝑘𝑘𝑉𝑉 − 1.3∠𝑗𝑗.5° 𝑘𝑘𝑉𝑉𝑽𝑽𝑅𝑅 = 1.𝑗6∠ − 14° 𝑘𝑘𝑉𝑉

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Power – Real Power & Power Factor20

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Power

The overall goal of a power distribution network is to transfer power from a source to loads

Instantaneous power: Power supplied by a source or absorbed by a load or

network element as a function of time

𝑝𝑝 𝑡𝑡 = 𝑣𝑣 𝑡𝑡 ⋅ 𝑖𝑖 𝑡𝑡 (8)

The nature of this instantaneous power flow is determined by the impedance of the load

Next, we’ll look at the instantaneous power delivered to loads of different impedances

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Instantaneous Power – Resistive Load

The voltage across the resistive load is

𝑣𝑣 𝑡𝑡 = 𝑉𝑉𝑝𝑝 cos 𝜔𝜔𝑡𝑡 + 𝛿𝛿

Current through the resistor is

𝑖𝑖 𝑡𝑡 =𝑉𝑉𝑝𝑝𝑅𝑅

cos 𝜔𝜔𝑡𝑡 + 𝛿𝛿

The instantaneous power absorbed by the resistor is

𝑝𝑝𝑅𝑅 𝑡𝑡 = 𝑣𝑣 𝑡𝑡 ⋅ 𝑖𝑖 𝑡𝑡 = 𝑉𝑉𝑝𝑝 cos 𝜔𝜔𝑡𝑡 + 𝛿𝛿 ⋅𝑉𝑉𝑝𝑝𝑅𝑅

cos 𝜔𝜔𝑡𝑡 + 𝛿𝛿

𝑝𝑝𝑅𝑅 𝑡𝑡 =𝑉𝑉𝑝𝑝2

𝑅𝑅cos2 𝜔𝜔𝑡𝑡 + 𝛿𝛿 =

𝑉𝑉𝑝𝑝2

𝑅𝑅12

1 + cos 𝑗𝜔𝜔𝑡𝑡 + 𝑗𝛿𝛿

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𝑝𝑝𝑅𝑅 𝑡𝑡 =𝑉𝑉𝑝𝑝2

𝑗𝑅𝑅1 + cos 𝑗𝜔𝜔𝑡𝑡 + 𝑗𝛿𝛿

Making use of the rms voltage

𝑝𝑝𝑅𝑅 𝑡𝑡 =2 𝑉𝑉𝑟𝑟𝑟𝑟𝑟𝑟

2

𝑗𝑅𝑅1 + cos 𝑗𝜔𝜔𝑡𝑡 + 𝑗𝛿𝛿

𝑝𝑝𝑅𝑅 𝑡𝑡 = 𝑉𝑉𝑟𝑟𝑟𝑟𝑟𝑟2

𝑅𝑅1 + cos 𝑗𝜔𝜔𝑡𝑡 + 𝑗𝛿𝛿 (9)

The instantaneous power absorbed by the resistor has a non-zero average value and a double-frequency component

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Power delivered to the resistive load has a non-zero average value and a double-frequency component

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Instantaneous Power – Capacitive Load

Now consider the power absorbed by a purely capacitive load Again, 𝑣𝑣 𝑡𝑡 = 𝑉𝑉𝑝𝑝 cos 𝜔𝜔𝑡𝑡 + 𝛿𝛿

The current flowing to the load is

𝑖𝑖 𝑡𝑡 = 𝐼𝐼𝑝𝑝 cos 𝜔𝜔𝑡𝑡 + 𝛿𝛿 + 90°where

𝐼𝐼𝑝𝑝 =𝑉𝑉𝑝𝑝𝑋𝑋𝐶𝐶

=𝑉𝑉𝑝𝑝

1/𝜔𝜔𝑗𝑗= 𝜔𝜔𝑗𝑗𝑉𝑉𝑝𝑝

The instantaneous power delivered to the capacitive load is

𝑝𝑝𝐶𝐶 𝑡𝑡 = 𝑣𝑣 𝑡𝑡 ⋅ 𝑖𝑖 𝑡𝑡𝑝𝑝𝐶𝐶 𝑡𝑡 = 𝑉𝑉𝑝𝑝 cos 𝜔𝜔𝑡𝑡 + 𝛿𝛿 ⋅ 𝜔𝜔𝑗𝑗𝑉𝑉𝑝𝑝 cos 𝜔𝜔𝑡𝑡 + 𝛿𝛿 + 90°

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Instantaneous Power – Capacitive Load

𝑝𝑝𝐶𝐶 𝑡𝑡 = 𝜔𝜔𝑗𝑗𝑉𝑉𝑝𝑝212

cos −90° + cos 𝑗𝜔𝜔𝑡𝑡 + 𝑗𝛿𝛿 + 90°

𝑝𝑝𝐶𝐶 𝑡𝑡 = 𝜔𝜔𝑗𝑗𝑉𝑉𝑝𝑝2

2⋅ cos 𝑗𝜔𝜔𝑡𝑡 + 𝑗𝛿𝛿 + 90°

In terms of rms voltage𝑝𝑝𝐶𝐶 𝑡𝑡 = 𝜔𝜔𝑗𝑗𝑉𝑉𝑟𝑟𝑟𝑟𝑟𝑟2 ⋅ cos 𝑗𝜔𝜔𝑡𝑡 + 𝑗𝛿𝛿 + 90°

This is a double frequency sinusoid, but, unlike for the resistive load, the average value is zero

(10)

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Instantaneous Power – Inductive Load

Now consider the power absorbed by a purely inductive load

Now the load current lags by 90°

𝑖𝑖 𝑡𝑡 = 𝐼𝐼𝑝𝑝 cos 𝜔𝜔𝑡𝑡 + 𝛿𝛿 − 90°where

𝐼𝐼𝑝𝑝 =𝑉𝑉𝑝𝑝𝑋𝑋𝐿𝐿

=𝑉𝑉𝑝𝑝𝜔𝜔𝐿𝐿

The instantaneous power delivered to the inductive load is

𝑝𝑝𝐿𝐿 𝑡𝑡 = 𝑣𝑣 𝑡𝑡 ⋅ 𝑖𝑖 𝑡𝑡

𝑝𝑝𝐿𝐿 𝑡𝑡 = 𝑉𝑉𝑝𝑝 cos 𝜔𝜔𝑡𝑡 + 𝛿𝛿 ⋅𝑉𝑉𝑝𝑝𝜔𝜔𝐿𝐿

cos 𝜔𝜔𝑡𝑡 + 𝛿𝛿 − 90°

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Instantaneous Power – Inductive Load

𝑝𝑝𝐿𝐿 𝑡𝑡 =𝑉𝑉𝑝𝑝2

𝜔𝜔𝐿𝐿12

cos 90° + cos 𝑗𝜔𝜔𝑡𝑡 + 𝑗𝛿𝛿 − 90°

𝑝𝑝𝐿𝐿 𝑡𝑡 =𝑉𝑉𝑝𝑝2

2𝜔𝜔𝐿𝐿⋅ cos 𝑗𝜔𝜔𝑡𝑡 + 𝑗𝛿𝛿 − 90°

In terms of rms voltage

𝑝𝑝𝐿𝐿 𝑡𝑡 =𝑉𝑉𝑟𝑟𝑟𝑟𝑟𝑟2

𝜔𝜔𝐿𝐿 ⋅ cos 𝑗𝜔𝜔𝑡𝑡 + 𝑗𝛿𝛿 − 90°

As for the capacitive load, this is a double frequency sinusoid with an average value of zero

(11)

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Instantaneous Power – General Impedance

Finally, consider the instantaneous power absorbed by a general RLC load

Phase angle of the current is determined by the angle of the impedance

𝑖𝑖 𝑡𝑡 = 𝐼𝐼𝑝𝑝 cos 𝜔𝜔𝑡𝑡 + 𝛽𝛽 The instantaneous power is

𝑝𝑝 𝑡𝑡 = 𝑉𝑉𝑝𝑝 cos 𝜔𝜔𝑡𝑡 + 𝛿𝛿 ⋅ 𝐼𝐼𝑝𝑝 cos 𝜔𝜔𝑡𝑡 + 𝛽𝛽

𝑝𝑝 𝑡𝑡 =𝑉𝑉𝑝𝑝𝐼𝐼𝑝𝑝

2cos 𝛿𝛿 − 𝛽𝛽 + cos 𝑗𝜔𝜔𝑡𝑡 + 𝛿𝛿 + 𝛽𝛽

𝑝𝑝 𝑡𝑡 = 𝑉𝑉𝑟𝑟𝑟𝑟𝑟𝑟𝐼𝐼𝑟𝑟𝑟𝑟𝑟𝑟 cos 𝛿𝛿 − 𝛽𝛽 + cos 𝑗𝜔𝜔𝑡𝑡 + 𝑗𝛿𝛿 − 𝛿𝛿 − 𝛽𝛽

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Using the following trig identity

cos 𝐴𝐴 − 𝐵𝐵 = cos 𝐴𝐴 cos 𝐵𝐵 + sin 𝐴𝐴 sin 𝐵𝐵

we get𝑝𝑝 𝑡𝑡 = 𝑉𝑉𝑟𝑟𝑟𝑟𝑟𝑟𝐼𝐼𝑟𝑟𝑟𝑟𝑟𝑟[cos 𝛿𝛿 − 𝛽𝛽 + cos 𝛿𝛿 − 𝛽𝛽 cos 𝑗𝜔𝜔𝑡𝑡 + 𝑗𝛿𝛿

+ sin 𝛿𝛿 − 𝛽𝛽 sin 𝑗𝜔𝜔𝑡𝑡 + 𝑗𝛿𝛿 ]

and𝑝𝑝 𝑡𝑡 = 𝑉𝑉𝑟𝑟𝑟𝑟𝑟𝑟𝐼𝐼𝑟𝑟𝑟𝑟𝑟𝑟 cos 𝛿𝛿 − 𝛽𝛽 1 + cos 𝑗𝜔𝜔𝑡𝑡 + 𝑗𝛿𝛿

+𝑉𝑉𝑟𝑟𝑟𝑟𝑟𝑟𝐼𝐼𝑟𝑟𝑟𝑟𝑟𝑟 sin 𝛿𝛿 − 𝛽𝛽 sin 𝑗𝜔𝜔𝑡𝑡 + 𝑗𝛿𝛿

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Letting𝐼𝐼𝑅𝑅 = 𝐼𝐼𝑟𝑟𝑟𝑟𝑟𝑟 cos 𝛿𝛿 − 𝛽𝛽 and 𝐼𝐼𝑋𝑋 = 𝐼𝐼𝑟𝑟𝑟𝑟𝑟𝑟 sin 𝛿𝛿 − 𝛽𝛽

we have𝑝𝑝 𝑡𝑡 = 𝑉𝑉𝑟𝑟𝑟𝑟𝑟𝑟𝐼𝐼𝑅𝑅 1 + cos 𝑗𝜔𝜔𝑡𝑡 + 𝑗𝛿𝛿

+𝑉𝑉𝑟𝑟𝑟𝑟𝑟𝑟𝐼𝐼𝑋𝑋 sin 𝑗𝜔𝜔𝑡𝑡 + 𝑗𝛿𝛿 (12)

There are two components to the power:

𝑝𝑝𝑅𝑅 𝑡𝑡 = 𝑉𝑉𝑟𝑟𝑟𝑟𝑟𝑟𝐼𝐼𝑅𝑅 1 + cos 𝑗𝜔𝜔𝑡𝑡 + 𝑗𝛿𝛿 (13)

is the power absorbed by the resistive component of the load, and

𝑝𝑝𝑋𝑋 𝑡𝑡 = 𝑉𝑉𝑟𝑟𝑟𝑟𝑟𝑟𝐼𝐼𝑋𝑋 sin 𝑗𝜔𝜔𝑡𝑡 + 𝑗𝛿𝛿 (14)

is the power absorbed by the reactive component of the load

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32

Real Power

According to (9) and (13), power delivered to a resistance has a non-zero average value Purely resistive load or a load with a resistive

component This is real power, average power, or active power

𝑃𝑃 = 𝑉𝑉𝑟𝑟𝑟𝑟𝑟𝑟𝐼𝐼𝑅𝑅

𝑃𝑃 = 𝑉𝑉𝑟𝑟𝑟𝑟𝑟𝑟𝐼𝐼𝑟𝑟𝑟𝑟𝑟𝑟 cos 𝛿𝛿 − 𝛽𝛽 (15)

Real power has units of watts (W) Real power is power that results in work (or heat

dissipation)

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33

Power Factor

The phase angle 𝛿𝛿 − 𝛽𝛽 represents the phase difference between the voltage and the current This is the power factor angle The angle of the load impedance

Note that the real power is a function of the cosine of the power factor angle

𝑃𝑃 = 𝑉𝑉𝑟𝑟𝑟𝑟𝑟𝑟𝐼𝐼𝑟𝑟𝑟𝑟𝑟𝑟 cos 𝛿𝛿 − 𝛽𝛽

This is the power factor

𝑝𝑝. 𝑓𝑓. = cos 𝛿𝛿 − 𝛽𝛽 (16)

For a purely resistive load, voltage and current are in phase

𝑝𝑝. 𝑓𝑓. = cos 𝛿𝛿 − 𝛽𝛽 = cos 0° = 1

𝑃𝑃 = 𝑉𝑉𝑟𝑟𝑟𝑟𝑟𝑟𝐼𝐼𝑟𝑟𝑟𝑟𝑟𝑟

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34

Power Factor

For a purely capacitive load, current leads the voltage by 90°

𝑝𝑝. 𝑓𝑓. = cos 𝛿𝛿 − 𝛽𝛽 = cos −90° = 0

𝑃𝑃 = 0

This is referred to as a leading power factor Power factor is leading for loads with capacitive reactance

For a purely inductive load, current lags the voltage by 90°

𝑝𝑝. 𝑓𝑓. = cos 𝛿𝛿 − 𝛽𝛽 = cos 90° = 0

𝑃𝑃 = 0

Loads with inductive reactance have lagging power factors Note that power factor is defined to always be positive

0 ≤ 𝑝𝑝. 𝑓𝑓.≤ 1

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Reactive & Complex Power35

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36

Reactive Power

The other part of instantaneous power, as given by (12), is the power delivered to the reactive component of the load

𝑝𝑝𝑋𝑋 𝑡𝑡 = 𝑉𝑉𝑟𝑟𝑟𝑟𝑟𝑟𝐼𝐼𝑟𝑟𝑟𝑟𝑟𝑟 sin 𝛿𝛿 − 𝛽𝛽 sin 𝑗𝜔𝜔𝑡𝑡 + 𝑗𝛿𝛿

Unlike real power, this component of power has zero average value

The amplitude is the reactive power

𝑄𝑄 = 𝑉𝑉𝑟𝑟𝑟𝑟𝑟𝑟𝐼𝐼𝑟𝑟𝑟𝑟𝑟𝑟 sin 𝛿𝛿 − 𝛽𝛽 𝑣𝑣𝑣𝑣𝑣𝑣

Units are volts-amperes reactive, or var Power that flows to and from the load reactance

Does not result in work or heat dissipation

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37

Complex Power

Complex power is defined as the product of the rms voltage phasor and conjugate rms current phasor

𝑺𝑺 = 𝑽𝑽𝑰𝑰∗ (18)

where the voltage has phase angle 𝛿𝛿

𝑽𝑽 = 𝑉𝑉𝑟𝑟𝑟𝑟𝑟𝑟∠𝛿𝛿

and the current has phase angle 𝛽𝛽

𝑰𝑰 = 𝐼𝐼𝑟𝑟𝑟𝑟𝑟𝑟∠𝛽𝛽 → 𝑰𝑰∗ = 𝐼𝐼𝑟𝑟𝑟𝑟𝑟𝑟∠ − 𝛽𝛽

The complex power is

𝑺𝑺 = 𝑽𝑽𝑰𝑰∗ = 𝑉𝑉𝑟𝑟𝑟𝑟𝑟𝑟∠𝛿𝛿 𝐼𝐼𝑟𝑟𝑟𝑟𝑟𝑟∠ − 𝛽𝛽

𝑺𝑺 = 𝑉𝑉𝑟𝑟𝑟𝑟𝑟𝑟𝐼𝐼𝑟𝑟𝑟𝑟𝑟𝑟∠ 𝛿𝛿 − 𝛽𝛽 (19)

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38

Complex Power

Complex power has units of volts-amperes (VA) The magnitude of complex power is apparent

power 𝑆𝑆 = 𝑉𝑉𝑟𝑟𝑟𝑟𝑟𝑟𝐼𝐼𝑟𝑟𝑟𝑟𝑟𝑟 𝑉𝑉𝐴𝐴 (20)

Apparent power also has units of volts-amperes Complex power is the vector sum of real power (in

phase with 𝑽𝑽) and reactive power (±90° out of phase with 𝑽𝑽)

𝑺𝑺 = 𝑃𝑃 + 𝑗𝑗𝑄𝑄 (21)

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39

Complex Power

Real power can be expressed in terms of complex power

𝑃𝑃 = 𝑅𝑅𝑒𝑒 𝑺𝑺

or in terms of apparent power

𝑃𝑃 = 𝑆𝑆 ⋅ cos 𝛿𝛿 − 𝛽𝛽 = 𝑆𝑆 ⋅ 𝑝𝑝. 𝑓𝑓.

Similarly, reactive power, is the imaginary part of complex power

𝑄𝑄 = 𝐼𝐼𝑚𝑚 𝑺𝑺

and can also be related to apparent power

𝑄𝑄 = 𝑆𝑆 ⋅ sin 𝛿𝛿 − 𝛽𝛽

And, power factor is the ratio between real power and apparent power

𝑝𝑝. 𝑓𝑓. = cos 𝛿𝛿 − 𝛽𝛽 =𝑃𝑃𝑆𝑆

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Passive Sign Convention40

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Power Convention – Load Convention

Applying a consistent sign convention allows us to easily determine whether network elements supply or absorb real and reactive power

Passive sign convention or load convention Positive current defined to enter the positive voltage

terminal of an element

If 𝑃𝑃 > 0 or 𝑄𝑄 > 0, then real or reactive power is absorbed by the element

If 𝑃𝑃 < 0 or 𝑄𝑄 < 0, then real or reactive power is supplied by the element

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42

Power Absorbed by Passive Elements

Complex power absorbed by a resistor

𝑺𝑺𝑹𝑹 = 𝑽𝑽𝑰𝑰𝑹𝑹∗ = 𝑉𝑉∠𝛿𝛿𝑉𝑉𝑅𝑅∠ − 𝛿𝛿

𝑺𝑺𝑹𝑹 =𝑉𝑉2

𝑅𝑅 Positive and purely real

Resistors absorb real power Reactive power is zero

Complex power absorbed by a capacitor 𝑺𝑺𝑪𝑪 = 𝑽𝑽𝑰𝑰𝑪𝑪∗ = 𝑉𝑉∠𝛿𝛿 −𝑗𝑗𝜔𝜔𝑗𝑗𝑉𝑉∠ − 𝛿𝛿𝑺𝑺𝑪𝑪 = −𝑗𝑗𝜔𝜔𝑗𝑗𝑉𝑉2

Negative and purely imaginary Capacitors supply reactive power Real power is zero

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43

Power Absorbed by Passive Elements

Complex power absorbed by an inductor

𝑺𝑺𝑳𝑳 = 𝑽𝑽𝑰𝑰𝑳𝑳∗ = 𝑉𝑉∠𝛿𝛿𝑉𝑉

−𝑗𝑗𝜔𝜔𝐿𝐿∠ − 𝛿𝛿

𝑺𝑺𝑳𝑳 = 𝑗𝑗𝑉𝑉2

𝜔𝜔𝐿𝐿 Positive and purely imaginary

Inductors absorb reactive power Real power is zero

In summary: Resistors absorb real power, zero reactive power Capacitors supply reactive power, zero real power Inductors absorb reactive power, zero real power

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Power Triangle44

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Power Triangle

Complex power is the vector sum of real power (in phase with 𝑽𝑽) and reactive power (±90° out of phase with 𝑽𝑽)

𝑺𝑺 = 𝑃𝑃 + 𝑗𝑗𝑄𝑄

Complex, real, and reactive powers can be represented graphically, as a power triangle

𝑄𝑄 = 𝑉𝑉𝐼𝐼 sin 𝛿𝛿 − 𝛽𝛽 var

𝑃𝑃 = 𝑉𝑉𝐼𝐼 cos 𝛿𝛿 − 𝛽𝛽 W

𝛿𝛿 − 𝛽𝛽

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46

Power Triangle

Quickly and graphically provides power information Power factor and power factor angle Leading or lagging power factor Reactive nature of the load – capacitive or inductive




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47

Lagging Power Factor

For loads with inductive reactance Impedance angle is positive Power factor angle is positive Power factor is lagging




𝑄𝑄 is positive The load absorbs reactive power

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48

Leading Power Factor

For loads with capacitive reactance Impedance angle is negative Power factor angle is negative Power factor is leading




𝑄𝑄 is negative The load supplies reactive power

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Power Factor Correction49

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50

Power Factor Correction

The overall goal of power distribution is to supply power to do work Real power

Reactive power does not perform work, but Must be supplied by the source Still flows over the lines

For a given amount of real power consumed by a load, we’d like to Reduce reactive power, 𝑄𝑄 Reduce 𝑆𝑆 relative to 𝑃𝑃, that is Reduce the p.f. angle, and Increase the p.f.

Power factor correction

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51

Power Factor Correction – Example

Consider a source driving an inductive load Determine:

Real power absorbed by the load Reactive power absorbed by the load p.f. angle and p.f.

Draw the power triangle

Current through the resistance is

𝑰𝑰𝑹𝑹 =120 𝑉𝑉

3 Ω= 40 𝐴𝐴

Current through the inductance is

𝑰𝑰𝑳𝑳 =120 𝑉𝑉𝑗𝑗𝑗 Ω

= 60∠ − 90° 𝐴𝐴

The total load current is𝑰𝑰 = 𝑰𝑰𝑹𝑹 + 𝑰𝑰𝑳𝑳 = 40 − 𝑗𝑗60 𝐴𝐴 = 72.𝑗∠ − 56.3° 𝐴𝐴

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52


The power factor angle is𝜃𝜃 = 𝛿𝛿 − 𝛽𝛽 = 0° − −56.3°𝜃𝜃 = 56.3°

The power factor is𝑝𝑝. 𝑓𝑓. = cos 𝜃𝜃 = cos 56.3°𝑝𝑝. 𝑓𝑓. = 0.55 lagging

Real power absorbed by the load is𝑃𝑃 = 𝑉𝑉𝐼𝐼 cos 𝜃𝜃 = 120 𝑉𝑉 ⋅ 72.1 𝐴𝐴 ⋅ 0.55𝑃𝑃 = 4.8 𝑘𝑘𝑘𝑘

Alternatively, recognizing that real power is power absorbed by the resistance

𝑃𝑃 = 𝑉𝑉𝐼𝐼𝑅𝑅 = 120 𝑉𝑉 ⋅ 40 𝐴𝐴 = 4.8 𝑘𝑘𝑘𝑘

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Reactive power absorbed by the load is

𝑄𝑄 = 𝑉𝑉𝐼𝐼 sin 𝜃𝜃 = 120 𝑉𝑉 ⋅ 72.1 𝐴𝐴 ⋅ 0.832𝑄𝑄 = 7.2 𝑘𝑘𝑣𝑣𝑣𝑣𝑣𝑣

This is also the power absorbed by the load inductance

𝑄𝑄 = 𝑉𝑉𝐼𝐼𝐿𝐿 = 120 𝑉𝑉 ⋅ 60 𝐴𝐴 = 7.2 𝑘𝑘𝑣𝑣𝑣𝑣𝑣𝑣

Apparent power is

𝑆𝑆 = 𝑉𝑉𝐼𝐼 = 120 𝑉𝑉 ⋅ 72.1 𝐴𝐴 = 8.65 𝑘𝑘𝑉𝑉𝐴𝐴

Or, alternatively

𝑆𝑆 = 𝑃𝑃2 + 𝑄𝑄2

𝑆𝑆 = 4.8 𝑘𝑘𝑘𝑘 2 + 7.2 𝑘𝑘𝑣𝑣𝑣𝑣𝑣𝑣 2 = 8.65 𝑘𝑘𝑉𝑉𝐴𝐴

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The power triangle: Here, the source is supplying

4.8 kW at a power factor of 0.55 lagging

Let’s say we want to reduce the apparent power supplied by the source

𝑄𝑄 = 7.2 kvar

𝑃𝑃 = 4.8 kW


Deliver 4.8 kW at a p.f. of 0.9 lagging

Add power factor correction

Add capacitors to supply reactive power

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For 𝑝𝑝. 𝑓𝑓. = 0.9, we need a power factor angle of

𝜃𝜃′ = cos−1 0.9 = 25.8°

Power factor correction will help flatten the power triangle:

𝑄𝑄 = 7.2 kvar

𝑃𝑃 = 4.8 kW

𝜃𝜃𝜃𝜃′

𝑄𝑄′

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56


The required reactive power absorbed (negative, so it is supplied) by the capacitors is

𝑄𝑄𝐶𝐶 = 𝑄𝑄′ − 𝑄𝑄 = 2.32 𝑘𝑘𝑣𝑣𝑣𝑣𝑣𝑣 − 7.2 𝑘𝑘𝑣𝑣𝑣𝑣𝑣𝑣

𝑄𝑄𝐶𝐶 = −4.88 𝑘𝑘𝑣𝑣𝑣𝑣𝑣𝑣

𝑄𝑄 = 7.2 kvar

𝑃𝑃 = 4.8 kW

𝜃𝜃𝜃𝜃′

𝑄𝑄′ = 2.32 kvar

Reactive power to the power-factor-corrected load is reduced from 𝑄𝑄 to 𝑄𝑄′

𝑄𝑄′ = 𝑃𝑃 tan 𝜃𝜃′

𝑄𝑄′ = 4.8 𝑘𝑘𝑘𝑘 ⋅ tan 25.8°

𝑄𝑄′ = 2.32 𝑘𝑘𝑣𝑣𝑣𝑣𝑣𝑣

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Reactive power absorbed by the capacitor is

𝑄𝑄𝐶𝐶 =𝑉𝑉2

𝑋𝑋𝐶𝐶 So the required capacitive reactance is

𝑋𝑋𝐶𝐶 =𝑉𝑉2

𝑄𝑄𝐶𝐶=

120 𝑉𝑉 2

−4.88 𝑘𝑘𝑣𝑣𝑣𝑣𝑣𝑣= −2.95 Ω

The addition of −𝑗𝑗𝑗.95 Ω provides the desired power factor correction

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Example Problems58

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The source voltage in the circuit is

𝑣𝑣 𝑡𝑡 = 2 ⋅ 𝑗𝑗0𝑉𝑉 cos 𝑗𝑗𝑗 ⋅ 60𝐻𝐻𝐻𝐻 ⋅ 𝑡𝑡 .

Determine the complex power delivered to the load.

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Two three-phase load are connected in parallel: 50 kVA at a power factor of 0.9, leading 125 kW at a power factor of 0.85, lagging.

Draw the power triangle and determine the combined power factor.

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Power is delivered to a single-phase load with an impedance of 𝑍𝑍𝐿𝐿 =3 + 𝑗𝑗𝑗 Ω at 120 V. Add power factor correction in parallel with the load to yield a power factor of 0.95, lagging.Determine the reactive power and impedance of the power factor correction component.

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Draw a phasor diagram for the following circuit. Draw a phasor for the voltage

across each component and for the current

Apply KVL graphically. That is, add the individual component phasors together graphically to show that the result is equal to the source voltage phasor.

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Balanced Three-Phase Networks67

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Balanced Three-Phase Networks

We are accustomed to single-phase power in our homes and offices A single line voltage referenced to a neutral

Electrical power is generated, transmitted, and largely consumed (by industrial customers) as three-phase power Three individual line voltages and (possibly) a neutral Line voltages all differ in phase by ±120°

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Δ- and Y-Connected Networks

Two possible three-phase configurations Applies to both sources and loads

Y-Connected Source 𝚫𝚫-Connected Source

Y-connected network has a neutral node Δ-connected network has no neutral

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70

Line-to-Neutral Voltages

A three-phase network is balanced if Sources are balanced The impedances connected to each phase are equal

In the Y network, voltages 𝑉𝑉𝑎𝑎𝐿𝐿, 𝑉𝑉𝑏𝑏𝐿𝐿, and 𝑉𝑉𝑐𝑐𝐿𝐿 are line-to-neutral voltages

A three-phase source is balanced if Line-to-neutral voltages have equal

magnitudes Line-to-neutral voltage are each 120°

out of phase with one another

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Line-to-Neutral Voltages

The line-to-neutral voltages are

𝑽𝑽𝒂𝒂𝒂𝒂 = 𝑉𝑉𝐿𝐿𝐿𝐿∠0°

𝑽𝑽𝒃𝒃𝒂𝒂 = 𝑉𝑉𝐿𝐿𝐿𝐿∠ − 120°

𝑽𝑽𝒄𝒄𝒂𝒂 = 𝑉𝑉𝐿𝐿𝐿𝐿∠ − 240° = 𝑉𝑉𝐿𝐿𝐿𝐿∠ + 120°

This is a positive-sequence or abc-sequencesource

𝑽𝑽𝒂𝒂𝒂𝒂 leads 𝑽𝑽𝒃𝒃𝒂𝒂, which leads 𝑽𝑽𝒄𝒄𝒂𝒂

Can also have a negative- or acb-sequencesource

𝑽𝑽𝒂𝒂𝒂𝒂 leads 𝑽𝑽𝒄𝒄𝒂𝒂, which leads 𝑽𝑽𝒃𝒃𝒂𝒂

We’ll always assume positive-sequence sources

Positive-Sequence Phasor Diagram:

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Line-to-Line Voltages The voltages between the three phases are line-to-

line voltages Apply KVL to relate line-to-line voltages to line-to-

neutral voltages𝑽𝑽𝒂𝒂𝒃𝒃 − 𝑽𝑽𝒂𝒂𝒂𝒂 + 𝑽𝑽𝒃𝒃𝒂𝒂 = 𝟎𝟎𝑽𝑽𝒂𝒂𝒃𝒃 = 𝑽𝑽𝒂𝒂𝒂𝒂 − 𝑽𝑽𝒃𝒃𝒂𝒂

We know that𝑽𝑽𝒂𝒂𝒂𝒂 = 𝑉𝑉𝐿𝐿𝐿𝐿∠0°

and𝑽𝑽𝒃𝒃𝒂𝒂 = 𝑉𝑉𝐿𝐿𝐿𝐿∠ − 120°

so 𝑽𝑽𝒂𝒂𝒃𝒃 = 𝑉𝑉𝐿𝐿𝐿𝐿∠0° − 𝑉𝑉𝐿𝐿𝐿𝐿∠ − 120° = 𝑉𝑉𝐿𝐿𝐿𝐿 𝑗∠0° − 𝑗∠ − 120°

𝑽𝑽𝒂𝒂𝒃𝒃 = 𝑉𝑉𝐿𝐿𝐿𝐿 1 − −12− 𝑗𝑗

32

= 𝑉𝑉𝐿𝐿𝐿𝐿32

+ 𝑗𝑗3

2

𝑽𝑽𝒂𝒂𝒃𝒃 = 3𝑉𝑉𝐿𝐿𝐿𝐿∠30°

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Line-to-Line Voltages

Again applying KVL, we can find 𝑽𝑽𝒃𝒃𝒄𝒄𝑽𝑽𝒃𝒃𝒄𝒄 = 𝑽𝑽𝒃𝒃𝒂𝒂 − 𝑽𝑽𝒄𝒄𝒂𝒂𝑽𝑽𝒃𝒃𝒄𝒄 = 𝑉𝑉𝐿𝐿𝐿𝐿∠ − 120° − 𝑉𝑉𝐿𝐿𝐿𝐿∠𝑗𝑗0°

𝑽𝑽𝒃𝒃𝒄𝒄 = 𝑉𝑉𝐿𝐿𝐿𝐿 −12− 𝑗𝑗

32

− −12

+ 𝑗𝑗3

2

𝑽𝑽𝒃𝒃𝒄𝒄 = 𝑉𝑉𝐿𝐿𝐿𝐿 −𝑗𝑗 3

𝑽𝑽𝒃𝒃𝒄𝒄 = 3𝑉𝑉𝐿𝐿𝐿𝐿∠ − 90°

Similarly,

𝑽𝑽𝒄𝒄𝒂𝒂 = 3𝑉𝑉𝐿𝐿𝐿𝐿∠𝑗50°

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Line-to-Line Voltages

The line-to-line voltages, with 𝑉𝑉𝑎𝑎𝐿𝐿 as the reference:

𝑽𝑽𝒂𝒂𝒃𝒃 = 3𝑉𝑉𝐿𝐿𝐿𝐿∠30°𝑽𝑽𝒃𝒃𝒄𝒄 = 3𝑉𝑉𝐿𝐿𝐿𝐿∠ − 90°𝑽𝑽𝒄𝒄𝒂𝒂 = 3𝑉𝑉𝐿𝐿𝐿𝐿∠𝑗50°

Line-to-line voltages are 3 times the line-to-neutral voltage

Can also express in terms of individual line-to-neutral voltages:

𝑽𝑽𝒂𝒂𝒃𝒃 = 3𝑽𝑽𝒂𝒂𝒂𝒂∠30°

𝑽𝑽𝒃𝒃𝒄𝒄 = 3𝑽𝑽𝒃𝒃𝒂𝒂∠30°

𝑽𝑽𝒄𝒄𝒂𝒂 = 3𝑽𝑽𝒄𝒄𝒂𝒂∠30°

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Currents in Three-Phase Networks75

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Line Currents in Balanced 3𝜙𝜙 Networks

𝑰𝑰𝒂𝒂 =𝑽𝑽𝑨𝑨𝑨𝑨𝑍𝑍𝑌𝑌

=𝑉𝑉𝐿𝐿𝐿𝐿∠0°𝑍𝑍𝑌𝑌

𝑰𝑰𝒃𝒃 =𝑽𝑽𝑩𝑩𝑨𝑨𝑍𝑍𝑌𝑌

=𝑉𝑉𝐿𝐿𝐿𝐿∠ − 120°

𝑍𝑍𝑌𝑌

𝑰𝑰𝒄𝒄 =𝑽𝑽𝑪𝑪𝑨𝑨𝑍𝑍𝑌𝑌

=𝑉𝑉𝐿𝐿𝐿𝐿∠ + 120°

𝑍𝑍𝑌𝑌

Line currents are balanced as long as the source and load are balanced

We can use the line-to-neutral voltages to determine the line currents Y-connected source and load Balanced load – all

impedances are equal: 𝑍𝑍𝑌𝑌

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Neutral Current in Balanced 3𝜙𝜙 Networks

Apply KCL to determine the neutral current

𝑰𝑰𝒂𝒂 = 𝑰𝑰𝒂𝒂 + 𝑰𝑰𝒃𝒃 + 𝑰𝑰𝒄𝒄

𝑰𝑰𝒂𝒂 =𝑉𝑉𝐿𝐿𝐿𝐿𝑍𝑍𝑌𝑌

𝑗∠0° + 𝑗∠ − 120° + 𝑗∠𝑗𝑗0°

𝑰𝑰𝒂𝒂 =𝑉𝑉𝐿𝐿𝐿𝐿𝑍𝑍𝑌𝑌

1 + −12− 𝑗𝑗

32

+ −12

+ 𝑗𝑗3

2

𝑰𝑰𝒂𝒂 = 0

The neutral conductor carries no current in a balanced three-phase network

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Y- and Δ-connected Loads78

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Three-Phase Load Configurations

As for sources, three-phase loads can also be connected in two different configurations

Y-Connected Load 𝚫𝚫-Connected Load

The Y load has a neutral connection, but the Δ load does not Currents in a Y-connected load are the line currents we just

determined Next, we’ll look at currents in a Δ-connected load

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Balanced Δ-Connected Loads

We can use line-to-line voltages to determine the currents in Δ-connected loads

𝑰𝑰𝑨𝑨𝑩𝑩 =𝑽𝑽𝑨𝑨𝑩𝑩𝑍𝑍Δ

=3𝑽𝑽𝑨𝑨𝑨𝑨∠30°

𝑍𝑍Δ=

3𝑉𝑉𝐿𝐿𝐿𝐿∠30°𝑍𝑍Δ

𝑰𝑰𝑩𝑩𝑪𝑪 =𝑽𝑽𝑩𝑩𝑪𝑪𝑍𝑍Δ

=3𝑽𝑽𝑩𝑩𝑨𝑨∠30°

𝑍𝑍Δ=

3𝑉𝑉𝐿𝐿𝐿𝐿∠ − 90°𝑍𝑍Δ

𝑰𝑰𝑪𝑪𝑨𝑨 =𝑽𝑽𝑪𝑪𝑨𝑨𝑍𝑍Δ

=3𝑽𝑽𝑪𝑪𝑨𝑨∠30°

𝑍𝑍Δ=

3𝑉𝑉𝐿𝐿𝐿𝐿∠150°𝑍𝑍Δ

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Balanced Δ-Connected Loads

Applying KCL, we can determine the line currents𝑰𝑰𝒂𝒂 = 𝑰𝑰𝑨𝑨𝑩𝑩 − 𝑰𝑰𝑪𝑪𝑨𝑨

𝑰𝑰𝒂𝒂 =3𝑉𝑉𝐿𝐿𝐿𝐿𝑍𝑍Δ

𝑗∠30° − 𝑗∠𝑗50°

𝑰𝑰𝒂𝒂 =3𝑉𝑉𝐿𝐿𝐿𝐿𝑍𝑍Δ

32

+ 𝑗𝑗12

− −3

2+ 𝑗𝑗

12

=3𝑉𝑉𝐿𝐿𝐿𝐿𝑍𝑍Δ

3 =3𝑉𝑉𝐿𝐿𝐿𝐿𝑍𝑍Δ

The other line currents can be found similarly:

𝑰𝑰𝒂𝒂 =3𝑉𝑉𝐿𝐿𝐿𝐿∠0°

𝑍𝑍Δ= 3𝑰𝑰𝑨𝑨𝑩𝑩∠ − 30°

𝑰𝑰𝒃𝒃 =3𝑉𝑉𝐿𝐿𝐿𝐿∠ − 120°

𝑍𝑍Δ= 3𝑰𝑰𝑩𝑩𝑪𝑪∠ − 30°

𝑰𝑰𝒄𝒄 =3𝑉𝑉𝐿𝐿𝐿𝐿∠𝑗𝑗0°

𝑍𝑍Δ= 3𝑰𝑰𝑪𝑪𝑨𝑨∠ − 30°

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Δ-Y Conversion82

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83

Δ − 𝑌𝑌 Conversion

Analysis is often simpler when dealing with 𝑌𝑌-connected loads Would like a way to convert Δ loads to 𝑌𝑌 loads (and vice

versa)

For a 𝑌𝑌 load and a Δ load to be equivalent, they must result in equal line currents

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Line currents for a 𝑌𝑌-connected load:

𝑰𝑰𝒂𝒂 =𝑉𝑉𝐿𝐿𝐿𝐿∠0°𝑍𝑍𝑌𝑌

𝑰𝑰𝒃𝒃 =𝑉𝑉𝐿𝐿𝐿𝐿∠ − 120°

𝑍𝑍𝑌𝑌

𝑰𝑰𝒄𝒄 =𝑉𝑉𝐿𝐿𝐿𝐿∠𝑗𝑗0°

𝑍𝑍𝑌𝑌

For a Δ-connected load:

𝑰𝑰𝒂𝒂 =3𝑉𝑉𝐿𝐿𝐿𝐿∠0°

𝑍𝑍Δ

𝑰𝑰𝒃𝒃 =3𝑉𝑉𝐿𝐿𝐿𝐿∠ − 120°

𝑍𝑍Δ

𝑰𝑰𝒄𝒄 =3𝑉𝑉𝐿𝐿𝐿𝐿∠𝑗𝑗0°

𝑍𝑍Δ

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Equating any of the three line currents, we can determine the impedance relationship

𝑉𝑉𝐿𝐿𝐿𝐿∠0°𝑍𝑍𝑌𝑌

=3𝑉𝑉𝐿𝐿𝐿𝐿∠0°

𝑍𝑍Δ

𝑍𝑍𝑌𝑌 = 𝑍𝑍Δ3

and 𝑍𝑍Δ = 3𝑍𝑍𝑌𝑌

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Per-Phase Analysis86

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Line-to-Neutral Schematics

For balanced networks, we can simplify our analysis by considering only a single phase A per-phase analysis Other phases are simply shifted by ±120°

For example, a balanced 𝑌𝑌-𝑌𝑌 circuit:

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One-Line Diagrams

Power systems are often depicted using one-line diagrams or single-line diagrams Not a schematic – not all wiring is shown

For example:

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Example Problems89

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Given the following balanced 3-𝜙𝜙 quantities:𝐕𝐕𝐵𝐵𝐶𝐶 = 480∠𝑗5° and 𝐈𝐈𝐵𝐵 = 𝑗𝑗∠ − 28°

Find:1) 𝐕𝐕𝐴𝐴𝐵𝐵2) 𝐕𝐕𝐴𝐴𝐿𝐿3) 𝐈𝐈𝐴𝐴4) 𝐈𝐈𝐶𝐶

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Find: Per-phase circuit

Line current, 𝐈𝐈𝐴𝐴 Load voltage

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Find: Per-phase circuit Line current, 𝐈𝐈𝐴𝐴 L-L and L-N load

voltages

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Power in Balanced 3𝜙𝜙 Networks95

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Instantaneous Power

We’ll first determine the instantaneous power supplied by the source Neglecting line impedance, this is also the power absorbed by the load

The phase 𝑣𝑣 line-to-neutral voltage is

𝑣𝑣𝑎𝑎𝐿𝐿 𝑡𝑡 = 2𝑉𝑉𝐿𝐿𝐿𝐿 cos 𝜔𝜔𝑡𝑡 + 𝛿𝛿 The phase 𝑣𝑣 current is

𝑖𝑖𝑎𝑎 𝑡𝑡 = 2𝐼𝐼𝐿𝐿 cos 𝜔𝜔𝑡𝑡 + 𝛽𝛽where 𝛽𝛽 depends on the load impedance

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Instantaneous Power

The instantaneous power delivered out of phase 𝑣𝑣of the source is

𝑝𝑝𝑎𝑎 𝑡𝑡 = 𝑣𝑣𝑎𝑎𝐿𝐿 𝑡𝑡 𝑖𝑖𝑎𝑎 𝑡𝑡

𝑝𝑝𝑎𝑎 𝑡𝑡 = 2𝑉𝑉𝐿𝐿𝐿𝐿𝐼𝐼𝐿𝐿 cos 𝜔𝜔𝑡𝑡 + 𝛿𝛿 cos 𝜔𝜔𝑡𝑡 + 𝛽𝛽

𝑝𝑝𝑎𝑎 𝑡𝑡 = 𝑉𝑉𝐿𝐿𝐿𝐿𝐼𝐼𝐿𝐿 cos 𝛿𝛿 − 𝛽𝛽 + 𝑉𝑉𝐿𝐿𝐿𝐿𝐼𝐼𝐿𝐿 cos 𝑗𝜔𝜔𝑡𝑡 + 𝛿𝛿 + 𝛽𝛽

The 𝑏𝑏 and 𝑐𝑐 phases are shifted by ±120° Power from each of these phases is

𝑝𝑝𝑏𝑏 𝑡𝑡 = 𝑉𝑉𝐿𝐿𝐿𝐿𝐼𝐼𝐿𝐿 cos 𝛿𝛿 − 𝛽𝛽 + 𝑉𝑉𝐿𝐿𝐿𝐿𝐼𝐼𝐿𝐿 cos 𝑗𝜔𝜔𝑡𝑡 + 𝛿𝛿 + 𝛽𝛽 − 240°

𝑝𝑝𝑐𝑐 𝑡𝑡 = 𝑉𝑉𝐿𝐿𝐿𝐿𝐼𝐼𝐿𝐿 cos 𝛿𝛿 − 𝛽𝛽 + 𝑉𝑉𝐿𝐿𝐿𝐿𝐼𝐼𝐿𝐿 cos 𝑗𝜔𝜔𝑡𝑡 + 𝛿𝛿 + 𝛽𝛽 + 240°

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Instantaneous Power

The total power delivered by the source is the sum of the power from each phase

𝑝𝑝3𝜙𝜙 𝑡𝑡 = 𝑝𝑝𝑎𝑎 𝑡𝑡 + 𝑝𝑝𝑏𝑏 𝑡𝑡 + 𝑝𝑝𝑐𝑐 𝑡𝑡

𝑝𝑝3𝜙𝜙 𝑡𝑡 = 3𝑉𝑉𝐿𝐿𝐿𝐿𝐼𝐼𝐿𝐿 cos 𝛿𝛿 − 𝛽𝛽+𝑉𝑉𝐿𝐿𝐿𝐿𝐼𝐼𝐿𝐿[cos 𝑗𝜔𝜔𝑡𝑡 + 𝛿𝛿 + 𝛽𝛽+ cos 𝑗𝜔𝜔𝑡𝑡 + 𝛿𝛿 + 𝛽𝛽 − 240°+ cos 𝑗𝜔𝜔𝑡𝑡 + 𝛿𝛿 + 𝛽𝛽 + 240° ]

Everything in the square brackets cancels, leaving

𝑝𝑝3𝜙𝜙 𝑡𝑡 = 3𝑉𝑉𝐿𝐿𝐿𝐿𝐼𝐼𝐿𝐿 cos 𝛿𝛿 − 𝛽𝛽 = 𝑃𝑃3𝜙𝜙

Power in a balanced 𝟑𝟑𝟑𝟑 network is constant In terms of line-to-line voltages, the power is

𝑃𝑃3𝜙𝜙 = 3𝑉𝑉𝐿𝐿𝐿𝐿𝐼𝐼𝐿𝐿 cos 𝛿𝛿 − 𝛽𝛽

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Complex Power

The complex power delivered by phase 𝑣𝑣 is

𝑺𝑺𝒂𝒂 = 𝑽𝑽𝒂𝒂𝒂𝒂𝑰𝑰𝒂𝒂∗ = 𝑉𝑉𝐿𝐿𝐿𝐿∠𝛿𝛿 𝐼𝐼𝐿𝐿∠𝛽𝛽 ∗

𝑺𝑺𝒂𝒂 = 𝑉𝑉𝐿𝐿𝐿𝐿𝐼𝐼𝐿𝐿∠ 𝛿𝛿 − 𝛽𝛽𝑺𝑺𝒂𝒂 = 𝑉𝑉𝐿𝐿𝐿𝐿𝐼𝐼𝐿𝐿 cos 𝛿𝛿 − 𝛽𝛽 + 𝑗𝑗𝑉𝑉𝐿𝐿𝐿𝐿𝐼𝐼𝐿𝐿 sin 𝛿𝛿 − 𝛽𝛽

For phase 𝑏𝑏, complex power is

𝑺𝑺𝒃𝒃 = 𝑽𝑽𝒃𝒃𝒂𝒂𝑰𝑰𝒃𝒃∗ = 𝑉𝑉𝐿𝐿𝐿𝐿∠ 𝛿𝛿 − 120° 𝐼𝐼𝐿𝐿∠ 𝛽𝛽 − 120° ∗

𝑺𝑺𝒃𝒃 = 𝑉𝑉𝐿𝐿𝐿𝐿𝐼𝐼𝐿𝐿∠ 𝛿𝛿 − 𝛽𝛽𝑺𝑺𝒃𝒃 = 𝑉𝑉𝐿𝐿𝐿𝐿𝐼𝐼𝐿𝐿 cos 𝛿𝛿 − 𝛽𝛽 + 𝑗𝑗𝑉𝑉𝐿𝐿𝐿𝐿𝐼𝐼𝐿𝐿 sin 𝛿𝛿 − 𝛽𝛽

This is equal to 𝑺𝑺𝒂𝒂 and also to phase 𝑺𝑺𝒄𝒄

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Complex Power

The total complex power is𝑺𝑺𝟑𝟑𝟑𝟑 = 𝑺𝑺𝒂𝒂 + 𝑺𝑺𝒃𝒃 + 𝑺𝑺𝒄𝒄𝑺𝑺𝟑𝟑𝟑𝟑 = 3𝑉𝑉𝐿𝐿𝐿𝐿𝐼𝐼𝐿𝐿∠ 𝛿𝛿 − 𝛽𝛽

𝑺𝑺𝟑𝟑𝟑𝟑 = 3𝑉𝑉𝐿𝐿𝐿𝐿𝐼𝐼𝐿𝐿 cos 𝛿𝛿 − 𝛽𝛽 + 𝑗𝑗3𝑉𝑉𝐿𝐿𝐿𝐿𝐼𝐼𝐿𝐿 sin 𝛿𝛿 − 𝛽𝛽

The apparent power is the magnitude of the complex power

𝑆𝑆3𝜙𝜙 = 3𝑉𝑉𝐿𝐿𝐿𝐿𝐼𝐼𝐿𝐿

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Complex Power

Complex power can be expressed in terms of the real and reactive power

𝑺𝑺𝟑𝟑𝟑𝟑 = 𝑃𝑃3𝜙𝜙 + 𝑗𝑗𝑄𝑄3𝜙𝜙

The real power, as we’ve already seen is

𝑃𝑃3𝜙𝜙 = 3𝑉𝑉𝐿𝐿𝐿𝐿𝐼𝐼𝐿𝐿 cos 𝛿𝛿 − 𝛽𝛽

The reactive power is

𝑄𝑄3𝜙𝜙 = 3𝑉𝑉𝐿𝐿𝐿𝐿𝐼𝐼𝐿𝐿 sin 𝛿𝛿 − 𝛽𝛽

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102

Advantages of Three-Phase Power

Advantages of three-phase power: For a given amount of power, half the amount of wire

required compared to single-phase No return current on neutral conductor

Constant real power Constant motor torque Less noise and vibration of machinery

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103

Three-Phase Power – Example

Determine Load voltage, 𝑽𝑽𝑨𝑨𝑩𝑩 Power triangle for the load Power factor at the load

We’ll do a per-phase analysis, so first convert the Δ load to a 𝑌𝑌 load

𝑍𝑍𝑌𝑌 =𝑍𝑍Δ3

= 1 + 𝑗𝑗0.5 Ω

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104


The per-phase circuit:

The line current is

𝑰𝑰𝑳𝑳 =𝑽𝑽𝒂𝒂𝒂𝒂

𝑍𝑍𝐿𝐿 + 𝑍𝑍𝑌𝑌=𝑗𝑗0∠0° 𝑉𝑉1.1 + 𝑗𝑗𝑗 Ω

=𝑗𝑗0∠0° 𝑉𝑉

1.45∠4𝑗.3° Ω

𝑰𝑰𝑳𝑳 = 80.7∠ − 42.3° 𝐴𝐴

The line-to-neutral voltage at the load is

𝑽𝑽𝑨𝑨𝑨𝑨 = 𝑰𝑰𝑳𝑳𝑍𝑍𝑌𝑌 = 80.7∠ − 42.3° 𝐴𝐴 1 + 𝑗𝑗0.5 Ω

𝑽𝑽𝑨𝑨𝑨𝑨 = 80.7∠ − 42.3° 𝐴𝐴 1.𝑗𝑗∠𝑗6.6° Ω

𝑽𝑽𝑨𝑨𝑨𝑨 = 90.25∠ − 15.71° V

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105


The line-to-line load voltage is𝑽𝑽𝑨𝑨𝑩𝑩 = 3𝑽𝑽𝑨𝑨𝑨𝑨∠30°

𝑽𝑽𝑨𝑨𝑩𝑩 = 𝑗56∠𝑗4.3° 𝑉𝑉

Alternatively, we could calculate line-to-line voltage from phase 𝐴𝐴 and phase 𝐵𝐵 line-to-neutral voltages

𝑽𝑽𝑨𝑨𝑩𝑩 = 𝑽𝑽𝑨𝑨𝑨𝑨 − 𝑽𝑽𝑩𝑩𝑨𝑨

𝑽𝑽𝑨𝑨𝑩𝑩 = 90.25∠ − 15.71° V − 90.25∠ − 135.71° V

𝑽𝑽𝑨𝑨𝑩𝑩 = 𝑗56∠𝑗4.3° 𝑉𝑉

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106


The complex power absorbed by the load is

𝑺𝑺𝟑𝟑𝟑𝟑 = 3𝑺𝑺𝑨𝑨 = 3𝑽𝑽𝑨𝑨𝑨𝑨𝑰𝑰𝑳𝑳∗

𝑺𝑺𝟑𝟑𝟑𝟑 = 3 90.𝑗5∠ − 15.71° 𝑉𝑉 80.7∠ − 42.3° 𝐴𝐴 ∗

𝑺𝑺𝟑𝟑𝟑𝟑 = 21.85 ∠𝑗6.6° 𝑘𝑘𝑉𝑉𝐴𝐴

𝑺𝑺𝟑𝟑𝟑𝟑 = 𝑗9.53 + 𝑗𝑗9.78 𝑘𝑘𝑉𝑉𝐴𝐴

The apparent power:𝑆𝑆3𝜙𝜙 = 21.85 𝑘𝑘𝑉𝑉𝐴𝐴

Real power:𝑃𝑃 = 𝑗9.53 𝑘𝑘𝑘𝑘

Reactive power:𝑄𝑄 = 9.78 𝑘𝑘𝑣𝑣𝑣𝑣𝑣𝑣

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The power triangle at the load:

𝑄𝑄 = 9.78 kvar

𝑃𝑃 = 𝑗9.53 kW

26.6°

The power factor at the load is

𝑝𝑝.𝑓𝑓. = cos 26.6° =𝑃𝑃𝑆𝑆

=𝑗9.53 𝑘𝑘𝑘𝑘21.85 𝑘𝑘𝑉𝑉𝐴𝐴

𝑝𝑝. 𝑓𝑓. = 0.89 𝑙𝑙𝑣𝑣𝑙𝑙𝑙𝑙𝑖𝑖𝑙𝑙𝑙𝑙

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Example Problems108

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Find: Source power Source power factor Load power Load power factor

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Find: Source power Load power Power lost in

lines

K. Webb ESE 470

section 2: three-phase power fundamentalsweb.engr.oregonstate.edu/~webbky/ese470_files/section...

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